Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const double PI = 3.14159265359;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long sy, a, b;
  cin >> sy >> a >> b;
  int n;
  cin >> n;
  vector<pair<long long, long long>> vsegs(n);
  for (auto &p : vsegs) {
    long long x, y;
    cin >> x >> y;
    p = {x, y};
  }
  int q;
  cin >> q;
  vector<pair<long long, long long>> vpts(q);
  for (auto &p : vpts) {
    long long x, y;
    cin >> x >> y;
    p = {x, y};
  }
  vector<long long> cs(n);
  cs[0] = vsegs[0].second - vsegs[0].first;
  for (int i = 1; i < n; i++) {
    cs[i] = cs[i - 1] + vsegs[i].second - vsegs[i].first;
  }
  for (int i = 0; i < q; i++) {
    double ans = 0;
    long long x1 = vpts[i].first, y1 = vpts[i].second;
    int lpos = 0, rpos = n - 1, l = 0, r = n;
    while (l < r) {
      int mid = (l + r) / 2;
      long long x2 = vsegs[mid].first;
      long long x3y1 = (x2 - x1) * (y1 - sy) + x1 * y1;
      if (x3y1 < a * y1)
        l = mid + 1;
      else
        r = mid;
    }
    lpos = l;
    l = 0, r = n;
    while (l < r) {
      int mid = (l + r) / 2;
      long long x2 = vsegs[mid].second;
      long long x3y1 = (x2 - x1) * (y1 - sy) + x1 * y1;
      if (x3y1 < b * y1)
        l = mid + 1;
      else
        r = mid;
    }
    rpos = l;
    double scaleFac = ((double)y1 - sy) / (y1);
    if (rpos > lpos) {
      ans += cs[rpos - 1];
      if (lpos) ans -= cs[lpos - 1];
      ans *= scaleFac;
    }
    int misl = lpos - 1, misr = rpos;
    if (misl != -1) {
      double x2 = vsegs[misl].second;
      double x3 = (double)(x1 - x2) * sy / y1 + x2;
      if (x3 > a) {
        ans += min((double)b, x3) - a;
      }
    }
    if (misr != n && misr != misl) {
      double x2 = vsegs[misr].first;
      double x3 = (double)(x1 - x2) * sy / y1 + x2;
      if (x3 < b) {
        ans += b - max(x3, (double)a);
      }
    }
    cout << setprecision(12) << ans << endl;
  }
  return 0;
}